I am trying not to use formulas..

\int_{}{}\ (^4\sqrt{3x+5}) (x) dx

Answer :  \frac{4}{81} (3x+5)^{\frac{5}{4} }(3x-4)+C

I am pretty sure I am on the right track but my weak algebra is making it harder



u= 3x+5
u'= 3

dx in tems of u =  \frac {1}{3}
and
x in tems of the new intgral =  (\frac{u-5}{3})


Here is the new integral!!!!!! I am going to simplify it now!!!!!!!
\frac {1}{3} \int_{}{} (\frac{u-5}{3}) u^{ \frac {1}{4} } du


Factoring out the 1/3
\frac {1}{3} \int_{}{} (\frac{1}{3})(u-5) u^{ \frac {1}{4} } du

\frac {1}{9} \int_{}{} (u-5) u^{ \frac {1}{4} } du


distribute the u^{1/4}, Notice that I can now directly integrate the function.

\frac {1}{9} ( \int_{}{} u ^{ \frac {5}{4} }du -5 \int_{}{} u ^{ \frac {1}{4} }du)


 \frac {1}{9} [ \frac {4}{9} u ^{ \frac {9}{4} }- 4u ^{ \frac {5}{4} } ]


factor out common crap

 \frac {1}{9} [ \frac {4}{9} u ^{ \frac {5}{4} }(u- 9 )] =  ( \frac {4}{81} u ^{ \frac {5}{4} })(u- 9 )

plug U back in

=  ( \frac {4}{81} (3x+5) ^{ \frac {5}{4} })(3x+5- 9 )

simplify and dont forget to add C!!!!!!!!!!

=  \frac {4}{81} (3x+5) ^{ \frac {5}{4} }(3x-4 )+C